Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 27 Apr 2015, 09:46

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In the figure above, points P and Q lie on the circle with

 Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
VP
Joined: 30 Sep 2004
Posts: 1490
Location: Germany
Followers: 4

Kudos [?]: 88 [3] , given: 0

In the figure above, points P and Q lie on the circle with [#permalink]  12 Sep 2005, 08:02
3
This post received
KUDOS
37
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

50% (01:43) correct 50% (00:42) wrong based on 846 sessions
Attachment:

image.JPG [ 3.91 KiB | Viewed 43353 times ]
In the figure above, points P and Q lie on the circle with center O. What is the value of s?

A. $$\frac{1}{2}$$

B. $$1$$

C. $$\sqrt{2}$$

D. $$\sqrt{3}$$

E. $$\frac{\sqrt{2}}{2}$$
[Reveal] Spoiler: OA

_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Intern
Joined: 23 Dec 2007
Posts: 1
Followers: 0

Kudos [?]: 58 [49] , given: 0

Re: gmatprep geo [#permalink]  23 Aug 2009, 14:54
49
This post received
KUDOS
9
This post was
BOOKMARKED
Check another approach for this problem.
Attachments

ps_ans1.JPG [ 25.21 KiB | Viewed 47612 times ]

Intern
Joined: 16 Jun 2009
Posts: 1
Followers: 1

Kudos [?]: 43 [35] , given: 0

Re: gmatprep geo [#permalink]  06 Sep 2009, 12:14
35
This post received
KUDOS
8
This post was
BOOKMARKED
There's a simple reason and misconception as to why people find this question confusing:
--> the 90 degree angle between P and Q seems like it is cut perfectly right down the centre >> to make two 45 degree angles.
--> that's why your brain instinctively thinks the answer is r = sq(3), since the distance seems the same from the left and to the right of the origin!

However, that's the trick to this question and that's why it is CRUCIAL that you train yourself to be highly critical and weary of accepting the simple answer!

Upon closer inspection of the question:

P(-sq(3),1)
Which means the triangle beneath point P has:
height = 1
base = sq(3)
meaning >> radius = 2 (30-60-90 triangle)

This follows the ideal case of the 30-60-90 triangle, which means that opposite to the height of 1, there is an angle of 30 degrees from the x-axis to point P. This means that from the y-axis to point P, there is an angle of 60 degrees (to make a 90 degrees total from x-axis to y-axis). Therefore on the right side, since the figure shows there is a 90 degree angle between point P and Q, 60 degrees have been taken up from the left of the y-axis, which leaves 30 degrees on the right of the y-axis. Furthermore, this leaves 60 degrees from Point Q to the positive x-axis. Therefore, making the exact same 30-60-90 triangle once again, since the radius is the same (the two points lie on the semi-circle). The 60 degree angle is now where the 30 degree angle was on the other triangle, which makes the height now sq(3) and base now 1 >> therefore, base r = 1!

**Have a look at the attachment and then it should make perfect sense!
Attachments

gmatprep-PSgeoQ.jpg [ 20.74 KiB | Viewed 47432 times ]

Current Student
Joined: 03 Aug 2006
Posts: 116
Location: Next to Google
Schools: Haas School of Business
Followers: 4

Kudos [?]: 142 [18] , given: 3

Re: coordinate geometry [#permalink]  16 Jun 2009, 09:18
18
This post received
KUDOS
2
This post was
BOOKMARKED
The answer is B.

Lets see how:
First draw a perpendicular from the x-axis to the point P. Lets call the point on the x-axis N. Now we have a right angle triangle $$\triangle NOP$$. We are given the co-ordinates of P as $$(-\sqrt{3}, 1)$$. i.e. $$ON=\sqrt{3}$$ and $$PN=1$$. Also we know angle $$PNO=90\textdegree$$. If you notice this is a $$30\textdegree-60\textdegree-90\textdegree$$triangle with sides $$1-sqrt3-2$$ and angle $$NOP=30\textdegree$$.

Similarly draw another perpendicular from the x-axis to point Q. Lets call the point on the x-axis R. Now we have a right angle triangle $$\triangle QOR$$. We know angle $$QRO=90\textdegree$$.

Now
angle NOP + angle POQ (= 90 -- given) + angle QOR = 180

$$\Rightarrow 30\textdegree+90\textdegree+angleQOR=180$$

$$\Rightarrow angleQOR=60$$

If you notice $$\triangle QOR$$ is also a $$30\textdegree-60\textdegree-90\textdegree$$triangle with sides $$1-sqrt3-2$$ and angle $$OQR=30\textdegree$$

The side opposite to this angle is OR = 1.

Hence answer is B.
Senior Manager
Joined: 04 May 2005
Posts: 284
Location: CA, USA
Followers: 1

Kudos [?]: 22 [10] , given: 0

[#permalink]  17 Sep 2005, 21:41
10
This post received
KUDOS
2
This post was
BOOKMARKED
there is no need for much calculation,
simply switch their X and Y

P and Q are always perpendicular and same distance from O

OP is perpendicular to OQ means you rotate OQ 90 degrees, you will
get OP, in terms of X,Y value, that is just to switch them and flip the sign
to have (-Y,X)
SVP
Joined: 04 May 2006
Posts: 1936
Schools: CBS, Kellogg
Followers: 19

Kudos [?]: 433 [6] , given: 1

Re: GMAT Prep 1 [#permalink]  27 Apr 2008, 21:49
6
This post received
KUDOS
1
This post was
BOOKMARKED
jbpayne wrote:
Thats what I thought, but the answer is B, 1. Anyone able to figure this one out?

There are more than one way to solve this. I like this way

The line PO has slope =-1/√3
The line QO has slope = t/s
PO and QO is perpendicular so [-1/√3]*t/s = -1 ==> t/s =√3 or √3/1 so s = 1
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5433
Location: Pune, India
Followers: 1325

Kudos [?]: 6720 [6] , given: 176

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  05 Jan 2011, 10:46
6
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
The method I suggested above is very intuitive and useful. I have been thinking of how to present it so that it doesn't look ominous. Let me try to present it in another way:

Imagine a clock face. Think of the minute hand on 10. The length of the minute hand is 2 cm. Its distance from the vertical and horizontal axis is shown in the diagram below. Let's say the minute hand moved to 1. Can you say something about the length of the dotted black and blue lines?
Attachment:

Ques1.jpg [ 7.87 KiB | Viewed 8985 times ]

Isn't it apparent that when the minute hand moved by 90 degrees, the green line became the black line and the red line became the blue line. So can I say that the black line $$= \sqrt{3}$$ cm and blue line = 1 cm

So if I imagine xy axis lying at the center of the clock, isn't it exactly like your question? The x and y co-ordinates just got inter changed. Of course, according to the quadrants, the signs of x and y coordinates will vary. If you did get it, tell me what are the x and y coordinates when the minute hand goes to 4 and when it goes to 7?
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Veritas Prep GMAT course is coming to India. Enroll in our weeklong Immersion Course that starts March 29!

Veritas Prep Reviews

SVP
Joined: 04 May 2006
Posts: 1936
Schools: CBS, Kellogg
Followers: 19

Kudos [?]: 433 [4] , given: 1

Re: GMAT Prep 1 [#permalink]  27 Apr 2008, 22:08
4
This post received
KUDOS
bsd_lover wrote:
Could you please elaborate the bolded part ?

sondenso wrote:
There are more than one way to solve this. I like this way

The line PO has slope =-1/√3
The line QO has slope = t/s
PO and QO is perpendicular so [-1/√3]*t/s = -1 ==> t/s =√3 or √3/1 so s = 1

I know I write like this [the colored] will raise the disscussion.
In general
slope of PO = a/b
slope of QO = t/s
PO and QO is perpendicular so [a/b]*[t/s]=-1, so t/s = -b/a
In this case, b = √3, a = -1 , so t/s = [-√3]/[-1] = √3/1, so s =1
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 27123
Followers: 4187

Kudos [?]: 40493 [4] , given: 5540

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  27 Apr 2010, 15:02
4
This post received
KUDOS
Expert's post
4
This post was
BOOKMARKED
zz0vlb wrote:
I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.

In the figure above, points P and Q lie on the circle with center O. What is the value of s?

Yes you are right.

Point Q in I quadrant $$(1, \sqrt{3})$$;
Point P in II quadrant $$(-\sqrt{3}, 1)$$;
Point T in III quadrant $$(-1, -\sqrt{3})$$ (OT perpendicular to OP);
Point R in IV quadrant $$(\sqrt{3},-1)$$ (OR perpendicular to OQ).
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5433
Location: Pune, India
Followers: 1325

Kudos [?]: 6720 [3] , given: 176

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  04 Jan 2011, 17:37
3
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
tonebreeze: As requested, I am giving you my take on this problem (though I guess you have already got some alternative solutions that worked for you)

I would suggest you first go through the explanation at the link given below for another question. If you understand that, solving this question is just a matter of seconds by looking at the diagram without using any formulas.
http://gmatclub.com/forum/coordinate-plane-90772.html#p807400

P is ($$-\sqrt{3}$$, 1). O is the center of the circle at (0,0). Also OP = OQ. When OP is turned 90 degrees to give OQ, the length of the x and y co-ordinates will get interchanged (both x and y co-ordinates will be positive in the first quadrant). Hence s, the x co-ordinate of Q will be 1 and y co-ordinate of Q will be $$\sqrt{3}$$.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Veritas Prep GMAT course is coming to India. Enroll in our weeklong Immersion Course that starts March 29!

Veritas Prep Reviews

Intern
Joined: 30 Aug 2005
Posts: 9
Followers: 0

Kudos [?]: 3 [2] , given: 0

[#permalink]  18 Sep 2005, 11:20
2
This post received
KUDOS
2
This post was
BOOKMARKED
Not sure if this is simpler or even different approach from the others(I maybe just formulizing qpoo's approach!)u.....

Lets say slope of OP =m1 and OQ =m2
Since OP perpendicular to OQ => m1m2=-1
Since m1= -1/sqrt(3)
=>m2=sqrt3
=>s=1,t=sqrt(3)
Manager
Joined: 08 Oct 2009
Posts: 67
Followers: 1

Kudos [?]: 20 [2] , given: 5

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  11 Oct 2009, 11:48
2
This post received
KUDOS
Since OP and OQ are the radii of the circle, and we know P has co-ordinates (-\sqrt{3}, 1), from the distance formula we have OP = OQ = 2. Therefore, s^2+t^2=4. -> (1)
Also from the right triangle, we know the length of PQ=2\sqrt{2}.
Again, using the distance formula, (s+\sqrt{3})^2 + (t-1)^2 = PQ^2 = 8. ->(2)

From (1) and (2) you can solve for s.
Hope this helps.
Current Student
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2801
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 191

Kudos [?]: 1112 [2] , given: 235

Re: Help on a Geometry PS Question [#permalink]  09 Sep 2010, 15:46
2
This post received
KUDOS
Attachment:

a.jpg [ 12.91 KiB | Viewed 46384 times ]

Line OP => y=mx => 1 = $$-sqrt{3} * m$$

=> slope of OP = $$\frac{-1}{sqrt{3}}$$

Since OP is perpendicular to OQ => m1*m2 = -1
=> slope of OQ = $$sqrt{3}$$

equation of line OQ => y = $$sqrt{3}x$$

since s and t lie on this line t= $$sqrt{3}s$$

also radius of circle = $$sqrt{ 1^2+ 3}$$ = 2 = $$sqrt{ s^2+ t^2}$$

=> $$4 = s^2+ t^2 = s^2 + 3s^2 = = 4s^2$$
=> s=1
----------------------------------------------------------------------

Attachment:

b.jpg [ 13.36 KiB | Viewed 46336 times ]

Using Similar triangles.
In triangle APO , angle APO + AOP = 90 ---------------------1
In triangle BOQ , angle BOQ + BQO= 90 ---------------------2

Also SINCE angle POQ = 90 as given.

angle AOP + angle BOQ = 90 ------------------3

Using equations 1 and 3 we get Angle APO = Angle BQO -----------4

using equation 2 ,3, 4
we get Angle APO = Angle BQO and AOP = OQB

Since hypotenuse is common => both the triangle are congruent

=> base of triangle APO = height of triangle AQO and vise versa.

Thus s = y co-ordinate of P = 1
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

Support GMAT Club by putting a GMAT Club badge on your blog/Facebook

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html

Last edited by gurpreetsingh on 09 Sep 2010, 16:00, edited 1 time in total.
Current Student
Status: Done with formalities.. and back..
Joined: 15 Sep 2012
Posts: 648
Location: India
Concentration: Strategy, General Management
Schools: Olin - Wash U - Class of 2015
WE: Information Technology (Computer Software)
Followers: 38

Kudos [?]: 400 [2] , given: 23

Re: Coordinate System and Geometry [#permalink]  19 Nov 2012, 23:55
2
This post received
KUDOS
agarboy wrote:
Can someone explain how to do this problem.. I have no clue how to even start on this.

In the attached picture, from question OA = $$\sqrt{3}$$
PA =1
Thus radius, OP = 2

Thus OPA is a 30, 60, 90 triangle. where angle POA is 30.
Also angle POQ is 90, given in question.

therefore, angle QOB = 60
And hence, angle OQB =30

trianlge QOB is also 30, 60, 90 triangle. in which we know OQ =OP =2 (radius)
Therefore BQ = $$\sqrt{3}$$
and OB = 1

We need value of r which is BQ = $$\sqrt{3}$$

Ans D it is!
Attachments

pqrs.jpg [ 10.93 KiB | Viewed 3459 times ]

_________________

Lets Kudos!!!
Black Friday Debrief

Intern
Joined: 28 Mar 2008
Posts: 35
Followers: 0

Kudos [?]: 2 [1] , given: 0

Re: GMAT Prep 1 [#permalink]  29 Apr 2008, 08:36
1
This post received
KUDOS
sondenso wrote:
bsd_lover wrote:
Could you please elaborate the bolded part ?

sondenso wrote:
There are more than one way to solve this. I like this way

The line PO has slope =-1/√3
The line QO has slope = t/s
PO and QO is perpendicular so [-1/√3]*t/s = -1 ==> t/s =√3 or √3/1 so s = 1

I know I write like this [the colored] will raise the disscussion.
In general
slope of PO = a/b
slope of QO = t/s
PO and QO is perpendicular so [a/b]*[t/s]=-1, so t/s = -b/a
In this case, b = √3, a = -1 , so t/s = [-√3]/[-1] = √3/1, so s =1

From the picture t^2 +s^2 = 4(radius = 2)
By substituting t =√3s we get s=1
Intern
Joined: 08 Nov 2009
Posts: 1
Followers: 0

Kudos [?]: 1 [1] , given: 1

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  09 Nov 2009, 19:25
1
This post received
KUDOS
Correct me if I'm wrong, (and forgive my english)
the coordinates of a point p(x,y) that lays on a circle can be expressed as p(sen,cos).
we know that sen π/6 = cos π/3,
and that sen π/3 = cos π/6

in this case I just have to invert x with y and y with x -> indeed we have p(y,x)

As we know x and y have to be positive we have to change the sign for x, the new coordinates are p (y,-x).

in this way to solve the problem doesn't take more than 20 secs.
hope I was enough clear.
Manager
Joined: 27 Feb 2010
Posts: 106
Location: Denver
Followers: 1

Kudos [?]: 156 [1] , given: 14

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  27 Apr 2010, 14:36
1
This post received
KUDOS
I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5433
Location: Pune, India
Followers: 1325

Kudos [?]: 6720 [1] , given: 176

Re: gmatprep geo [#permalink]  20 Oct 2013, 19:35
1
This post received
KUDOS
Expert's post
obs23 wrote:
chetanojha wrote:
Check another approach for this problem.

Can we assume that coordinates of the center O are (0,0) here, can we always assume so?

From the figure it is clear that O lies at (0, 0). O is the point where x axis and y axis intersect in the figure.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Veritas Prep GMAT course is coming to India. Enroll in our weeklong Immersion Course that starts March 29!

Veritas Prep Reviews

Manager
Joined: 12 Oct 2009
Posts: 115
Followers: 2

Kudos [?]: 30 [0], given: 3

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  13 Oct 2009, 04:29
DenisSh wrote:
Sorry, didn't get that part:
badgerboy wrote:
Again, using the distance formula, (s+\sqrt{3})^2 + (t-1)^2 = PQ^2 = 8. ->(2)

The method used to find the length of OP or OQ from origin i.e. the distance formula has been used here as well to find the length of PQ with P as (-[square_root]3,1) and Q as (s,t) which gives us
{s - (-[square_root]3)}^2 + {t-1}^2 = PQ^2 = 8 ->equation 2

from Right angle Triangle POQ we had got the length of PQ^2 as 8
Senior Manager
Affiliations: PMP
Joined: 13 Oct 2009
Posts: 312
Followers: 3

Kudos [?]: 110 [0], given: 37

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  13 Oct 2009, 12:15
Yes Answer is 1.

S^2 + T^2 = 4 --> equation 1

(t-1)^2 + (s-(-sqrt(3))^2 = 2^2 + 2^2 = 8 --> equation 2

solving for s gives 1
_________________

Thanks, Sri
-------------------------------
keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

Re: Plane Geometry, Semicircle from GMATPrep   [#permalink] 13 Oct 2009, 12:15

Go to page    1   2   3    Next  [ 48 posts ]

Similar topics Replies Last post
Similar
Topics:
1 In the figure above, points P and Q lie on the circle with 1 29 Jan 2012, 11:41
In the figure above, points P and Q 1 24 Dec 2008, 15:45
1 In the figure below, points P and Q lie on the circle with 15 25 Oct 2007, 10:13
In the figure, point P and Q lie on the circle with center 11 03 Sep 2007, 08:07
In the attached figure, points P and Q lie on the circle 8 15 Jun 2006, 23:55
Display posts from previous: Sort by

# In the figure above, points P and Q lie on the circle with

 Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.